What volume of air at N.T.P containing 21 % oxygen by volume is required to completely burn 1000 g of sulphur containing 4 % incombustible matter ?

Answer

`3360 L`

Explanation

Amount of incombustible matter in sulphur `= (1000 xx 4)/(100) = 40 g`
Amount of pure matter `= 1000 - 40 = 960 g`
Step I. Volume of oxygen required
The chemical equation for the reaction is
`underset(32g)(1//8S_(B))+underset(22.4L)(O_(2))rarrSO_(2)`
`32g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P.` `= 22.4 L`
`960g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P = ((22.4 L) xx (960g))/((32g)) = 972 L`
Step II. Volume air required
Volume of oxygen required `= 972 L`
Volume of air required `= (672 L) xx (100)/(200) = 3360 L`